Percentages and percentage problems

Percentages are the most widely used mathematics in everyday life: discounts, loans, savings, taxes, pay rises, exam results — all are measured in per cent. In the exam percentage problems appear in almost every paper and often come with a “trap”: successive changes, the wrong base, compound interest. This lesson starts from what a percentage means and covers the three basic problems, successive changes, compound interest and mixture problems — with the reasons behind each rule. The topic is closely linked with ratio and proportion, fractions and the exponential function.

§1. What a per cent is and the three basic problems

A per cent is one hundredth: 1% = 1/100. So p% = p/100; 25% = 0.25 = 1/4, 150% = 1.5. To turn a percentage into a decimal, move the point two places left; to go back, move it right.
All percentage problems reduce to three basic types:

These three formulas are three forms of one proportion: “a is 100%, b is p%”. You find whichever quantity is unknown.
Useful equivalents: 10% = 1/10, 20% = 1/5, 25% = 1/4, 50% = 1/2, 12.5% = 1/8, 33⅓% = 1/3. They speed up mental arithmetic: 25% of 240 is just 240 : 4 = 60.

Worked example (from the official sample)

A 2.4 m by 4 m carpet lies in a 4 m × 6 m room. What percentage of the room’s area does the carpet cover?

1. Carpet area 2.4·4 = 9.6 m², room area 4·6 = 24 m².
2. The whole (100%) is the room’s area, since the question asks for “a percentage of the room’s area”.
3. 9.6/24·100% = 40%.

§2. Increase and decrease

Why multiply? Adding p% of a to a gives a + a·p/100 = a(1 + p/100). The number in brackets is the multiplier: +15% → 1.15, −30% → 0.7.
Successive changes multiply, they do not add, because each new percentage is taken of the new value:

• +20%, then −20%: 1.2·0.8 = 0.96 — the result is 4% less than the start.
• +10% twice: 1.1² = 1.21 → +21%, not 20%.

Worked example

A price was 50 000 som, rose by 20% and then fell by 10%. Final price and overall change?

1. 50 000·1.2 = 60 000.
2. 60 000·0.9 = 54 000 som.
3. The product of multipliers 1.2·0.9 = 1.08 — an overall change of +8%.

Worked example

In a workshop the number of workers rose by 20% and each worker’s productivity by 25%. By what percentage did output rise?

1. Output = number of workers × productivity, so the multipliers are multiplied.
2. 1.2·1.25 = 1.5.
3. Output rose by 50% (not 20% + 25% = 45%).

§3. Reverse problems: finding the original value

A common trap: “after a 20% discount the price is 48 000 som. What was the original price?” Many add 20% to 48 000 and get 57 600 — wrong, because the 20% was taken of the original price, not the new one.

Worked example

After a 20% discount an item costs 48 000 som. What was its original price?

1. Multiplier: 1 − 0.2 = 0.8.
2. x·0.8 = 48 000.
3. x = 48 000 : 0.8 = 60 000 som.

§4. “How many per cent more / less” — choosing the base

A percentage always depends on which number it is taken of. In “A is ... more than B” the base is B (the number being compared with).

• A is 25% more than B: A = 1.25B.
• Then B compared with A: B/A = 1/1.25 = 0.8 → 20% less.

Distinguish percentage points from per cent: if a rate rises from 10% to 12%, that is a rise of 2 percentage points but a relative rise of 20%.

§5. Compound interest

If p% is added each period and the next percentage is calculated on the new amount (bank savings, population growth):

With compound interest the amount grows like a geometric progression, with simple interest like an arithmetic one. So over a long term compound interest gives far more.

Worked example

1 000 000 som is invested for 2 years at 10% a year compound interest.

1. End of year one: 1 000 000·1.1 = 1 100 000.
2. End of year two: 1 100 000·1.1 = 1 210 000 som.
3. Simple interest would give 1 200 000 — the 10 000 difference is “interest on interest”.

§6. Solution and mixture problems

The concentration of a solution is the mass of the substance divided by the mass of the solution:

Key idea: find the quantity that does not change. When water is added or evaporates, the mass of the substance stays the same; only the mass of the solution changes. When solutions are mixed, the masses of substance add up. When fruit is dried, water leaves but the dry matter stays.

Worked example

50 g of water is added to 200 g of a 15% salt solution. The new concentration?

1. Salt: 200·0.15 = 30 g — it does not change.
2. New solution: 200 + 50 = 250 g.
3. 30/250 = 0.12 → 12%.

Worked example

Fresh fruit contains 80% water, dried fruit 20%. How many kilograms of fresh fruit give 10 kg of dried fruit?

1. Dry matter in the dried fruit: 10·0.8 = 8 kg.
2. In fresh fruit dry matter is 20%.
3. 8 : 0.2 = 40 kg.

Check yourself

1. Find 15% of 240.

2. 30% of which number is 45?

3. A is 25% more than B. By what percentage is B less than A?

4. A price rose by 10% twice in a row. The overall percentage rise?

5. 300 g of a 40% solution is mixed with 200 g of a 10% solution. Find the concentration.

6. A price fell from 80 000 to 68 000. By what percentage?

7. After a 25% rise a price is 50 000. The original price?

8. 2 000 000 som is invested for 3 years at 12% simple interest. The final amount?

Common mistakes

• Adding successive percentages: +20% and −20% give −4%, not 0%.
• Using the wrong base — the change is divided by the old value.
• Adding the percentage to the new value in a reverse problem — divide by the multiplier instead.
• Confusing compound and simple interest.
• Changing the mass of the substance when water is added to a solution.
• Mixing up percentage points and per cent (relative change).